i have string like below
#a b#mail.com has joined
i want to show #a b c as mention, but hard to detect space, so i put the string like below before using regex
#``a b#mail.com`` has joined
So i want to detect start with
#``
and end with
``
can anyone help me the regex, i tried so much but still not working, here is the regex im testing
^#``.*``{3,}
You might use a capture group and capture any char except an # using a negated character class.
^#``([^#\r\n]+).*?``
The pattern matches
^#`` Start of string, match # and 2 backticks
( Capture group 1
[^#\r\n]+ Match 1+ occurrences of any char except # or a newline
) Close group 1
.*?`` Match as least as possible chars until 2 backticks
Regex demo
Related
I wanted to know the regex expression that detects names starting with #. For eg, in the sentence "Hi #Steve Rogers, how are you?", I want to extract out #Steve Rogers using regex. I tried using Pattern.compile("#\\s*(\\w+)").matcher(text), but only "#Steve" get detected. What else should I use.??
Thanks
Try (#[\w\s]+)
It will only capture word and spaces after the #
See example at https://regex101.com/r/4Pv9bu/1
If you don't want to match an # sign followed by a space only like # and if there can be more than a single word after it:
(?<!\S)#\w+(?:\h+\w+)?
Explanation
(?<!\S) Assert a whitespace boundary to the left
# Match literally
\w+ Match 1+ word characters
(?:\s+\w+)? Optionally match 1+ horizontal whitespace chars and 1+ word chars
Regex demo
In Java
String regex = "(?<!\\S)#\\w+(?:\\h+\\w+)?";
I am looking for a regx for below expression:
my $text = "1170 KB/s (244475 bytes in 2.204s)"; # I want to retrieve last ‘2.204’ from this String.
$text =~ m/\d+[^\d*](\d+)/; #Regexp
my $num = $1;
print " $num ";
Output:
204
But I need 2.204 as output, please correct me.
Can any one help me out?
The regex is doing exactly what you asked it to: It is matching digits \d+, followed by one non-digit or star [^\d*], followed by digits \d+. The only thing that matches that in your string is 204.
If you want a quick fix, you can just move the parentheses:
m/(\d+[^\d*]\d+)/
This would (with the above input) match what you want. A more exact way to put it would be:
m/(\d+\.\d+)/
Of course this will match any float precision number, so if you can have more of those, that's not a good idea. You can shore it up by using an anchor, like so:
m/(\d+\.\d+)s\)/
Where s\) forces the match to occur at only that place. Further strictures:
m/\(\d+\D+(\d+\.\d+)s\)/
You might also want to account for the possibility of your target number not being a float:
m/\(\d+\D+(\d+\.?\d*)s\)/
By using ? and * we allow for those parts not to match at all. This is not recommended to do unless you are using anchors. You can also replace everything in the capture group with [\d.]+.
If you are not fond of matching the parentheses, you can match the text:
m/bytes in ([\d.]+)s/
I'd go with the second marker as indicator where you are in the string:
my ($num) = ($text =~ /(\d+\.\d+)s/);
with explanations:
/( # start of matching group
\d+ # first digits
\. # a literal '.', take \D if you want non-numbers
\d+ # second digits
)/x # close the matching group and the regex
You had the matching groups wrong. Also the [^\d] is a bit excessive, generally you can negate some of the backspaced special classes (\d,\h, \s and \w) with their respective uppercase letter.
Try this regex:
$text =~ m/\d+[^\d]*(\d+\.?\d*)s/;
That should match 1+ digits, a decimal point if there is one, 0 or more decimal places, and make sure it's followed by a "s".
Here is a piece of code
while($l=~/(\\\s*)$/) {
statements;
}
$l contains a line of text taken form file, in effect this code is for go through lines in file.
Questions:
I don't clearly understand what the condition in while is doing. I think it is trying to match group of \ followed by some number of white spaces at the end of line and loop should stop whenever a line ends with \ and may be some white spaces. I am not sure of it.
I came across statement $a ~= s/^(.*$)/$1/ . What I understand that ^ will force matching at the beginning of string, but in (.*$) would mean match all the characters at the end of string . Dose it mean that the statement is trying to find if any group of character at the end is same as group of character in the beginning of text ?
It is interesting to note that this statement:
while ( $l =~ /(\\\s*)$/ ) {
Is an infinite loop unless $l is altered inside the loop so that the regex no longer matches. As has already been mentioned by others, this is what it matches:
( ... ) a capture group, captures string to $1 (that's the number one, not lower case L)
\\ matches a literal backslash
\s* matches 0 or more whitespace characters.
$ matches end of line with optional newline.
Since you do not have the /g modifier, this regex will not iterate through matches, it will simply check if there is a match, resetting the regex each iteration, thereby causing an endless loop.
The statement
$a ~= s/^(.*$)/$1/
Looks rather pointless. It captures a string of characters up until end of string, then replaces it with itself. The captured text is stored in $1 and is simply replaced. The only marginally useful thing about this regex is that:
It matches up until newline \n, and nothing further, which may be of some use to a parser. A period . matches any character except newline, unless the /s modifier is present on the regex.
It captures the line in $1 for future use. However, a simple /^(.*$)/ would do the same.
1. the while
Usually while (regex) is used with the /g modifier, otherwise, if it matches, you get an infinite loop (unless you exit the loop, like using last).
statements would be executed continuously in an infinite loop.
In your case, adding the g
while($l=~/(\\\s*)$/g)
will have the while make only one loop, due to the $ - making a match unique (whatever matches up to the end of string is unique, as $ marks the end, and there is nothing after...).
2. $a ~= s/^(.*$)/$1/
This is a substitution. If the string ^.*$ matches (and it will, since ^.*$ matches (almost, see comment) anything) it is replaced with... $1 or what's inside the (), ie itself, since the match occurs from 1st char to the end of string
^ means beginning of string
(.*) means all chars
$ end of string
so that will replace $a with itself - probably not what you want.
it matches a literal backslash followed by 0 or more spaces followed by the end of the line.
it executes statements for all the lines in that text file that contain a \, followed by zero or more spaces ( \s* ), at the end of the line ($).
It matches lines that end with a backslash character, ignoring any trailing whitespace characters.
Ending a line with a backslash is used in some languages and data files to indicate that the line is being continued on the next line. So I suspect this is part of a parser that merges these continuation lines.
If you enter a regular expression at RegExr and hover your mouse over the pieces, it displays the meaning of each piece in a tooltip.
(\\\s*)$ this regex means --- a \ followed by zero or more number of white space characters which is followed by end of the line. Since you have your regex in (...), you can extract what you matched using $1, if you need.
http://rubular.com/r/dtHtEPh5DX
EDIT -- based on your update
$a ~= s/^(.$)/$1/ --- this is search and replace. So your regex matches a line which contains exactly one character (since you use . http://www.regular-expressions.info/dot.html), except a new-line character. Since you use (...), the character which matched the regex is extracted and stored in variable a
EDIT -- you changed your regex so here is the updated answer
$a ~= s/^(.*$)/$1/ -- same as above except now it matches zero or more characters (except new-line)
I would like to translate all instances of a character with two characters. The usual way I would do it is:
$text =~ s/a/aa/g;
I only want single instances of a character to be doubled. So aa would remain aa and not turn into aaaa.
I am thinking I have to use variables in the s/// statement but I cannot find any suitable pattern here or on the net.
Match instances of a that are not next to another a:
s/(?<!a)a(?!a)/aa/g;
I've to match a regular-expression, stored in a variable:
#!/bin/env perl
use warnings;
use strict;
my $expr = qr/\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)/sx;
$str = "abcd[3] xyzg[4:0]";
if ($str =~ m/$expr/) {
print "\n%%%%%%%%% $`-----$&-----$'\n";
}
else {
print "\n********* NOT MATCHED\n";
}
But I'm getting the outout in $& as
%%%%%%%%% -----abcd[3] xyzg-----[4:0]
But expecting, it shouldn't go inside the if clause.
What is intended is:
if $str = "abcd xyzg" => %%%%%%%%% -----abcd xyzg----- (CORRECT)
if $str = "abcd[2] xyzg" => %%%%%%%%% -----abcd[2] xyzg----- (CORRECT)
if $str = "abcd[2] xyzg[3] => %%%%%%%%% -----abcd[2] xyzg[3]----- (CORRECT)
if $str = "abcd[2:0] xyzg[3] => ********* NOT MATCHED (CORRECT)
if $str = "abcd[2:0] xyzg[3:0] => ********* NOT MATCHED (CORRECT)
if $str = "abcd[2] xyzg[3:0]" => ********* NOT MATCHED (CORRECT/INTENDED)
but output is %%%%%%%%% -----abcd[2] xyzg-----[3:0] (WRONG)
OR better to say this is not intended.
In this case, it should/my_expectation go to the else block.
Even I don't know, why $& take a portion of the string (abcd[2] xyzg), and $' having [3:0]?
HOW?
It should match the full, not a part like the above. If it didn't, it shouldn't go to the if clause.
Can anyone please help me to change my $expr pattern, so that I can have what is intended?
By default, Perl regexes only look for a matching substring of the given string. In order to force comparison against the entire string, you need to indicate that the regex begins at the beginning of the string and ends at the end by using ^ and $:
my $expr = qr/^\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)$/;
(Also, there's no reason to have the /x modifier, as your regex doesn't include any literal whitespace or # characters, and there's no reason for the /s modifier, as you're not using ..)
EDIT: If you don't want the regex to match against the entire string, but you want it to reject anything in which the matching portion is followed by something like "[0:0]", the simplest way would be to use lookahead:
my $expr = qr/^\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\]|(?=[^[\w])|$ ))/x;
This will match anything that takes the following form:
beginning of the string (which your example in the comments seems to imply you want)
zero or more whitespace characters
one or more word characters
optional: [, one or more digits, ]
one or more whitespace characters
one or more word characters
one of the following, in descending order of preference:
[, one or more digits, ]
an empty string followed by (but not including!) a character that is neither [ nor a word character (The exclusion of word characters is to keep the regex engine from succeeding on "a[0] bc[1:2]" by only matching "a[0] b".)
end of string (A space is needed after the $ to keep it from merging with the following ) to form the name of a special variable, and this entails the reintroduction of the /x option.)
Do you have any more unstated requirements that need to be satisfied?
The short answer is your regexp is wrong.
We can't fix it for you without you explaining what you need exactly, and the community is not going to write a regexp exactly for your purpose because that's just too localized a question that only helps you this one time.
You need to ask something more general about regexps that we can explain to you, that will help you fix your regexp, and help others fix theirs.
Here's my general answer when you're having trouble testing your regexp. Use a regexp tool, like the regex buddy one.
So I'm going to give a specific answer about what you're overlooking here:
Let's make this example smaller:
Your pattern is a(bc+d)?. It will match: abcd abccd etc. While it will not match bcd nor bzd in the case of abzd it will match as matching only a because the whole group of bc+d is optional. Similarly it will match abcbcd as a dropping the whole optional group that couldn't be matched (at the second b).
Regexps will match as much of the string as they can and return a true match when they can match something and have satisfied the entire pattern. If you make something optional, they will leave it out when they have to including it only when it's present and matches.
Here's what you tried:
qr/\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)/sx
First, s and x aren't needed modifiers here.
Second, this regex can match:
Any or no whitespace followed by
a word of at least one alpha character followed by
optionally a grouped square bracketed number with at least one digit (eg [0] or [9999]) followed by
at least one white space followed by
a word of at least one alpha character followed by
optionally a square bracketed number with at least one digit.
Clearly when you ask it to match abcd[0] xyzg[0:4] the colon ends the \d+ pattern but doesn't satisfy the \] so it backtracks the whole group, and then happily finds the group was optional. So by not matching the last optional group, your pattern has matched successfully.